2520 is the smallest number that is divisible by all the numbers
from 1 to 10.

How do you find such a number? It doesn't need to be
1×2×3×4×5×6×7×8×9×10, there are many smaller
answers. To consider a simpler example, 12 is the smallest number
divisible by 1, 2, 3 and 4 but it is smaller than 1×2×3×4. There
are different ways to find the answer, all of which amount essentially
to looking at the prime factorizations of all the numbers and keeping
the highest exponents of each prime that occur. The simplest way to
describe the answer is as follows: Keep the highest power of each
prime, and throw away all the rest. For example, in the numbers 1
through 10, the highest prime powers are 8=23, 9=32, 5=51 and
7=71. All the other numbers are composite (e.g. 6=2×3 and
10=2×5) or smaller powers of the primes already mentioned (2=21,
4=22 and 3=31).

This number is also related to 10 in another way: it is 10 times the
central number (252) on row 10 of
Pascal's Triangle. This is sort of related, because 252 =
10!/(5!×5!); see 5040 and 3628800.

3003 is also a palindrome, and the 77thtriangular
number. It shares this property with 1, 3, 6, 55, 66, 171, 595,
666, and many others (Sloane's A3098).
Note that the index (77) is itself a palindrome. Possibly because of
the simple formula Tn = n(n+1)/2, there are also at least two
cases with a repunit index: T1111=617716 and
T111111=6172882716.

This is 6×7×8×9, which also happens to be equal to the sum
5×6×7×8 + 4×5×6×7 + 3×4×5×6 + 2×3×4×5 +
1×2×3×4. A little investigation shows that this is one of a
family of similar patterns: 1+2=3, 1×2+2×3+3×4=4×5,
1×2×3+2×3×4+3×4×5+4×5×6=5×6×7, and so on.
The sums/products for the sequence: 1, 3, 20, 210, 3024, 55440,
1235520, 32432400, ... (Sloane's A6963) which counts "planar
embedded trees of N nodes".

The solution to the "Monkey and Coconuts problem" (1926 Saturday
Evening Post version), which according to Martin Gardner62,63
is "probably the most worked on and least often solved of all"
Diophantine equations. Here is the problem (in my words):

Five sailors, stranded on a desert island, spend the day
gathering coconuts and then go to sleep, agreeing to divide them up
in the morning.
After a while, one sailor wakes up, concerned that he might
not get his fair share. He gives one coconut to the monkey to keep
him quiet, then divides the pile into five parts, and finds
that it divides evenly. He hides one fifth and puts the other 4/5 back
into one pile (the monkey keeps his).
One after another, each of the other four sailors does the
same thing  wakes up, gives a coconut to the monkey, divides the
rest into five parts, which comes out evenly, hides one-fifth and
puts the remaining 4/5 back into one pile.
In the morning, the men together divide the remaining
pile and find it divides equally into five parts. How many coconuts
were there in the beginning?

A little trial-and-error reveals that the answer is probably somewhat
large. One can work it out methodically by working backwards from the
end: The final pile F must be a multiple of both 4 and 5, so it can
be any number in the sequence {20, 40, 60, 80, ...}, the pile prior to
that has to have been 5/4F+1, which admits the numbers in the
sequence {76, 176, 276, ...}, etc. There are also more elaborate
solution techniques, some of which are outlined in an excellent
Numberphile video. There is more than one answer, but
the smallest is 3121. The general solution for n sailors is
nn-n+1 for odd n, and (2n-1)nn-n+1 for even
n.64

The number of seconds in an hour (60×60), the number of
fingers in a cord (units of length), the number of shekels in a talent
(units of weight), the number of years in the (Babylonian) long
"saros", and the product of squares of the simplest
Pythagorean triangle (32×42×52). All of these were
significant to the Babylonians.

Although it is not a divisibility record-setter
itself, 3600 is the square of the popular record-setter (and base of
the Sumerian/Babylonian number system) 60.

The division of the hour into 60 minutes or 3600 seconds is the most
familiar relic of the old Sumerian base-60 numbering system. It also
survived in the divisions of angles into degrees, minutes (short for
"minute divisions"), seconds, thirds14, and so on. The division of
an hour into 3600 parts also happens to be convenient and useful. The
hour had been long established as 1/12 part of the daylight
period (to such an extent that, in many cultures, the length of the
hour increased and decreased with the seasons!). The pace represented
by 3600 beats per hour arises naturally because it is close to the
frequency of a human heartbeat; most people feel their heartbeat when
resting quietly. Since hours and heartbeats were already pretty well
established, it was useful to choose a number that was pretty close to
the right ratio but also was arithmetically convenient to work with.
3600 was by far the best choice.

4879 is a Kaprekar number under the most relaxed
Kaprekar rules, in which the point of division can be in a place other
than the number of digits in the original number. 4879^2 is 23804641,
but instead of dividing it into 2380+4641, it is divided into
238+04641 = 4879. Even though the original number 4879 has four digits
and its square has eight, the square is divided unevenly
giving three digits to the "left half" and five digits to the "right
half" 04681. Under the more strict Kaprekar rules, the division always
happens in the place corresponding to the number of digits in the
original number.

In 1875 Edouard Lucas considered the problem: "A certain number
of cannonballs can be arranged in a square on the ground, and can also
be stacked into a square pyramid. How many are there?"

The number of cannonballs in the pyramid is the sum of consecutive
squares 12+22+32+...+x2.
This problem is formally stated as the diophantine equation
x(x+1)(2x+1)=6y2. The only known answers were the trivial
case x=y=1 and the intended answer to the problem, x=24,
y=70, with y2=4900 cannonballs.

Lucas conjectured that there were no other solutions; it was proven
by Watson in 1918, but using a rather complicated method. Interest in
the original problem continued, and in 1985 D.G. Ma published the
first proof using entirely elementary means.

Amazingly, the fact that 24(24+1)(2×24+1)=702 is connected to
26-dimensional Lorentzian space-time, the
Leech lattice, and the
Monster group. The connection appears96 to be
evident in Conway and Sloane [146] and is made more explicit
by Borcherds in his 1984 Ph.D. thesis [164].

Mike Hill, instructor of an abstract algebra course at University of
Virginia, describes it this way97 in course notes on the Monster
group:

A path on which the time-distance is always zero in a higher
dimensional (> 4) space-time (Lorentzian space) yields a perpendicular
Euclidean space of two dimensions lower.
ex. 26-dimensional Lorentzian space yields the 24-dimensional
Euclidean space which contains the Leech lattice.

The Leech lattice contains a point (0,1,2,3,4,...,24,70)
Time distance from origin in Lorentzian space
0 = 02 + 12 + 22 + ... + 232 + 242 - 702
This point lies on a light ray through the origin

Borcherd said a string moving in space-time is only nonzero if
space-time is 26-dimensional.

5041 = 7!+1 = 712. This is the highest known case of a
square which is one more than a factorial. The
other cases are 25 = 4!+1 = 52 and 121 = 5!+1 = 112. There are
several similar special properties of numbers (for examples, see
39, 89, 91 and 51381)
where the distribution falls off so quickly that it's difficult to see
if there are only a finite number of numbers with the property. In
this case for example, the odds of N! + 1 being a square are about 1
in √N!, assuming there is no special relationship between the
distribution of squares and factorials. Since the factorials grow very
quickly, the infinite sum

SUM [ 1 / √N! ]

converges very quickly, and in fact it's a bit of a surprise that
there are as many as three solutions for N! + 1 = M2. The fact
that there are three suggests that the distribution of factorials and
squares might have a relationship  but one should be wary of the
Strong Law of Small Numbers (again see 91).

An approximation of the number of years in the "Mayan long count"
calendar, according to some theories. This calendar began at a date
that is equivalent to 3114 BC in the Gregorian system; it counts days
and has an integer value expressed in a 5-digit mixed-base system
using bases 13, 20, 20, 18 and 20 in each of the five places. The
rightmost two places, counting in cycles of 18×20, correspond to a
360-day year, and the other three places count "years"
(periods of 360 solar days). The total number of days is
1872000. Expressed more accurately, the "5126 years" is
13×202×18×20/365.242189670 = 5125.3662719.
See 2012.

The use of 13 as the base of the highest place in this calendar's
counting system is uncertain, and it could just as likely be 20, which
would lead to a "long count" of 18×204/365.242189670 or about
7885 years.

Number of feet in a mile. 5280 is close to the Roman
mile defined as 1000 paces, or 5000 feet, but was adapted to
accommodate other units of length including the furlong (660 feet)
and chain (66 feet). These numbers (66, 660 and 5280)
are all multiples of 11 because of their relation to the old
unit of length called a rod, also called a perch or
pole, which is the same length as an old farmer's tool called an "ox
goad"92, used in plughing. However, the origin appears93 to go
back to Denmark and Prussia sometime before the 11th century;
evidence for this fact lies in the fact that the English rod agrees
with the old Danish fod and the Prussian Rheinfuss to within less
than half an inch, or 0.19% of the total length.

Number of mean solar days in the
metonic cycle of 19 tropical years or
235 synodic months. Its integer approximation (6940)
can be used as the basis of a lunisolar calendar that repeats every 19
years, and with some effort you can also make the months fairly
regular. For example, every year can have 12 months each of which has
a specific name, 6 with 29 days and 6 with 30 days; 7 out of 19 years
have a 13th month of 30 days, and 4 of those 7 years also add an extra
day to one of the normal 29-day months. Other similar systems are
possible but all solutions have equal complexity; actual lunisolar
calendars (like the Hebrew calendar) are more complex
but achieve greater accuracy.

Although it is complex, a lunisolar calendar has very strong
practical motivations. The importance of the solar day
is obvious; the tropical year is important to anyone
living in a climate with seasons. And the
synodic month tells us when it is possible to see at
night by the light of the moon, and what time of day the high and low
tides will take place. These things are important even in modern urban
society. Under a lunisolar calendar you can agree to meet outdoors at
8PM every month on the 15th of the month and know that there will be
moonlight (weather permitting), or you can agree to go fishing every
10th of the month at 5AM and know there will be a low tide (assuming
that's how the tide lines up at your location).

7980 = 15×19×28, the product of three numbers that have an
important relation to calendars in the Roman, Byzantine and Christian
worlds. 28 years is the "solar cycle", 4×7, the number of
years it takes before any date falls on the same day of the week again
(that's 4 years per leap-year cycle, and 7 days per week). 19 is the
length of the metonic cycle, see 19 for more. 15
years is the "indiction cycle", a period related to certain phenomena
such as taxation. Since all three numbers are relatively prime to each
other, the least common multiple is 7980. The "Julian day number",
used in astronomy, is based on a system proposed in 1583 just after
the adoption of the Gregorian calendar. Julian day 1 is January 1st
4713 BC. The year 4713 BC happened to be the most recent time that all
three cycles (the 15- 18- and 28-year cycles) were aligned. It is
convenient primarily because it pre-dates all recorded history, even
in China, Egypt, Greece and Mesopotamia.

(Personal: For a while during my childhood the numbers 7,
27 and 127 were my favorite 1- 2- and 3-digit numbers.
I have since forgotten why the properties of 127 appealed to me, but I
suspect it was partly because it ends in "27". If I had continued the
series, I probably would have picked 8127, because it is divisible by
7 and by 27, and consists of the cubes 8, 1 and 27 strung together;
also the "81" is 3×27.)

The smallest number that is not a palindrome, and is divisible by
the number you get when you reverse its digits: 8712 = 4×2178. The
next number like this is 9801=992. The numbers with this property
comprise Sloane's A31877 and their factors are
A8919. (Thanks to Tavi Laiu for this one). See
also 1089. Another related property: 8712 × 2178 =
664.

The necessary letters are combined to make a quantity, so for example
the number 8127 would be ͵ΗΡΚΖ. For numbers above this they used a
capital letter mu (Μ) to represent 10000, whose name in Greek is
myriad. To distinguish it from Μ=40, a small alpha α was
written above the Μ. The precise way in which larger numbers were
handled differed through the centuries in ancient Greece. In the most
ambitious system, a number (like ͵ΗΡΚΖ) was written in small letters
(in this case ͵ηρκζ) directly above a letter Μ to represent 10000
raised to that power; see 1040000. However, in much
more common usage the small number placed above the letter Μ merely
multiplies the value; see 108.

10001 is a "generalised Fermat number", of the form a2n +
1, with a=10 and n=2. They include the normal
Fermat numbers (when a=2 or 4), OEIS sequence
A78303 when a=6, A152581 when a=8, A80176,
A178426, A152587, etc. If a is odd they are even (and
thus composite); when a is even they have some of the properties of
the normal Fermat numbers, such as being prime a little
more often than one might expect, or having reasonably large prime
factors. For example, 10001 is 73×137, and
1016+1 = 353×449×641×1409×69857. The "generalised
Fermat primes" would be all primes expressible as a2n + 1,
which is the same as OEIS sequence A2496, primes of form n2
+ 1.

10080 = 60 × 24 × 7, the number of minutes in a week. 10080 is
also a divisibility record-setter with 72 divisors. Thus,
there are 72 different ways to divide a week into equal parts that are
whole multiples of minutes. 10080 is also 2×7!, see
40320 and 604800.

In France just after the revolution they switched to the
now-standard system of weights and measures based on powers of 10 (SI,
systeme internationale, or the "metric system"). They also set up
(for a while) a calendar involving a 10-day week, and dividing the day
into powers of 10. It didn't last long, and France switched back to
the Gregorian calendar a few years later.

But there have been other times since then that people have
considered the idea of measuring time in powers of 10. Clearly this
"metric time" idea has some appeal, because there are dozens of web
sites about it (do a search for "metric time"). Perhaps best-known is
Swatch's "Internet time", based on the division of the day into 1000
parts called "beats" (with 86.4 seconds per "beat"). For subdivisions
of the day it is easy, the hours and minutes we have now are just
arbitrary divisions and there is no conflict with the other important
environmental cycles.

When considering a metric replacement for the
synodic month and tropical year, clearly
there is no good practical solution. These are all very important
physical cycles and have strange ratios that cannot be changed (at
least not yet :-) and cannot easily be adapted to work with powers of
10.

Personally, I think the existing standard time system is just as
cool as any 10-based system would be, because of the utilitarian
properties of the factor record-setters and cute things like the
minutes in February and the
86400000 property. However, if you insist on setting up
a "metric time" system, you could do worse than to take advantage of
the fact that a minute and a week are almost at an exact ratio of
10000. Such a system would leave the definition of "week"
unchanged and define a special "minute" which is exactly 7/10000 of a
day (which comes out to exactly 60.48 "standard" seconds). Then all
weekly events would always start at the same minute-mark, which would
be a 4-digit number from 0000 to 9999. You'd still have to deal with
the fact that the number of weeks per year is odd (the fact that it's
close to 100/2 makes this a little more bearable), the number of
minutes per day is odd, and there is no suitable "month".

In 2011 November,
this was identified
as the "smallest uninteresting number" in an episode of QI, based
on its being the smallest (positive) integer not appearing in the
listed terms of any sequence in the OEIS. It is
important to note that some sequences, such as A0027, contain
every integer  so this property of 12407 was only possible because
each entry in the OEIS gives only a few lines of initial terms for
each sequence. There is a lot more about this in [205].

The smaller powers of 11 (121, 1331 and 14641) give rows of
Pascal's Triangle. Pascal's Triangle is a rather useful table of
numbers (called binomial coefficients) that is formed by
continually adding numbers in a cascade starting from a single 1, as
shown here:

row 0:

1

row 1:

1

1

row 2:

1

2

1

row 3:

1

3

3

1

row 4:

1

4

6

4

1

row 5:

1

5

10

10

5

1

row 6:

1

6

15

20

15

6

1

row 7:

1

7

21

35

35

21

7

1

row 8:

1

8

28

56

70

56

28

8

1

row 9:

1

9

36

84

126

126

84

36

9

1

etc

etc

etc

Each is the sum of the (one or) two numbers above it.

These numbers have many applications because they count
"combinations", such as "how many combinations of 3 colors can you
make if you have 5 colors to choose from?" (the answer is the 3rd
number in row 5 of the triangle: 10). Another question with the same
answer is, "If you flip a coin 5 times, how many ways are there to end
up getting heads exactly 3 times?" (the answer is also 10).

A general formula that gives the value of item N in row M is:

NCM = M!/((M-N)!×N!)

For example, 3C5 = 5!/((5-3)!×3!) = 120/(2×6) = 10. However
this formula gets rather impractical to use when M is large, even
when the value being computed is pretty reasonable. For example,
consider 3C143=143!/(140!×3!). 143! has 248 digits, and
140!×3! has 242 digits  so you probably couldn't even get the
answer on a calculator, as the calculator would overflow. However,
most of the 143! in the numerator cancels out with the 140! in the
denominator, leaving 3C143 = 143×142×141/3! = 477191, which
you could even calculate by hand if you needed to.